For reference: In the drawing, $T$ is the point of tangency, $LN || AT$, $OH = 4$ and $LN^2+AM^2=164$. Calculate $HN$. (Answer: $8$)
*Both circles have the same radius.
Progress:
By Stewart's theorem, $\triangle LNT-NH \implies LN^2+NT^2=2(HN^2+HT^2)$, but $HT=OH=4$, therefore $LN^2+NT^2=2(HN^2+16)...$.
Comment
With given values $NH\neq 8$ and if $NH=8$ , $OH=4$ and $LN^2+AM^2=164$ then the figure can not be constructed. What I could construct is:
-Radius of circles $r=5.5$
-$LN=8.94$, $AM=8.9$ $\Rightarrow LN^2+AM^2=158$
-$LH=OH=4$
which gives : $HN^2=8,9^2-4^2=8$
-$\angle AMN=90^o$
-$NM=10.5$
-These measures give $AN=13.8$
You have to show these measure by calculation.